Conical pendulum in rotating frame

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SUMMARY

The discussion centers on the dynamics of a conical pendulum at the North Pole, analyzing its behavior in a rotating frame due to Earth's rotation. The pendulum's period when moving east, denoted as T_E, is longer than when moving west, T_W, due to the Coriolis effect and the need to "catch up" with the rotating ground. The equation of motion incorporates the Coriolis force and a modified gravitational acceleration, represented as mg'. Understanding these forces is crucial for calculating the pendulum's period accurately.

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  • Understanding of conical pendulum dynamics
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  • Knowledge of rotational motion and forces
  • Ability to analyze equations of motion in rotating frames
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  • Learn about modified gravitational acceleration in rotating frames
  • Explore free body diagram techniques for analyzing forces
  • Investigate the differences between rotating and non-rotating frames of reference
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Physics students, educators, and researchers interested in classical mechanics, particularly those studying pendulum dynamics and the effects of Earth's rotation on motion.

StrangelyQuarky
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Homework Statement


A pendulum of length l at the north pole is moving in a circle to the east at an angle \theta to the vertical. It has some period T_E as measured in the rotating Earth frame. The experiment is then repeated except now the pendulum is moving to the west with period T_W. The question asks which period is longer, and to calculate the relative time difference in the periods.

Homework Equations



In the rotating frame, the equation of motion of the pendulum involves the Coriolis force and a modified gravitational acceleration due to the rotation of the Earth: m\mathbf{a'} = m\mathbf{g'} + \mathbf{T} -2m(\mathbf{\omega '} \times \mathbf{v'})

where T is the tension in the wire.

The Attempt at a Solution


Since the Earth rotates counterclockwise as viewed looking down on the north pole, the pendulum that moves to the east has a longer period because the Earth's rotation means it has to "catch up" with the rotating ground (I think?). However, I am at a loss as to how to calculate the period from equation of motion.
 
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StrangelyQuarky said:
In the rotating frame, the equation of motion of the pendulum involves the Coriolis force and a modified gravitational acceleration due to the rotation of the Earth: m\mathbf{a'} = m\mathbf{g'} + \mathbf{T} -2m(\mathbf{\omega '} \times \mathbf{v'})
Can you describe in more detail the "modified gravitational acceleration" ##\mathbf{g'}##? What is its magnitude and direction?

What does the symbol ##\mathbf{\omega '}## stand for?

What about the "centrifugal force" in the rotating frame? Or does your ##\mathbf{g'}## include that?

Since the Earth rotates counterclockwise as viewed looking down on the north pole, the pendulum that moves to the east has a longer period because the Earth's rotation means it has to "catch up" with the rotating ground (I think?).
Yes, the "catching up" is from the point of view of an observer in a nonrotating. In fact, you can work the problem by analyzing it from the nonrotating frame. But, if you want to work it out in the rotating frame, then follow the usual recipe of starting with a free body diagram, etc.
 
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